TERMS AND SEQUENCES

[margie]

$1000 is invested at 7% per annum, compound interest. After the interest was credited at the end of each year, $100 was always withdrawn. The account balance at the start of each year is given by:

A tn=(1.07)n tn-1-100
B tn=(1.07)tn-1-100
C tn= (1.07)n (tn-1-100)
D tn=(1.07) (tn-1-100)
E tn= (0.07)n tn-1-100

 

[tkhunny]

E - No good. It's only the interest, not the accumulation.
A, C - No good. What's that exponent doing in there. Getting from n to (n+1) is oney one full year.
D - No good. Why would we calculate interest credits for the $100 we took out?

Please, PLEASE figure out a little better notation. See that "tex" button? It can do things like this: \(\displaystyle t_{n}=(1.07 \cdot t_{n-1}) - 100\)

 

[margie]

Thanks.. I didnt see the tex button. Thanks for telling me about it and for putting me on the right track with the question.
Margie

 

[Deleted member 4993]

or you can do that with sub button too!

t[sub:2l3yaga5]n[/sub:2l3yaga5] = 1.07 * t[sub:2l3yaga5]n-1[/sub:2l3yaga5] - 100

 

[soroban]

Hello, Margie!

$1000 is invested at 7% per annum, compound interest.
After the interest was credited at the end of each year, $100 was always withdrawn.
The account balance at the start of each year is given by:

. . \(\displaystyle \begin{array}{ccccc}A) & t_n &=& (1.07)^n t_{n-1}-100 \\ \\[-3mm] B) &t_n &=& (1.07)t_{n-1}-100 \\ \\[-3mm] C) & t_n &=& (1.07)^n(t_{n-1}-100) \\ \\ [-3mm] D) & t_n &=& (1.07)(t_{n-1}-100) \\ \\[-3mm] E) & t_n &=& (0.07)^nt_{n-1}-100 \end{array}\)

At the beginning of a year, the balance was \(\displaystyle t_{n-1}\)

During that year, it earns 7% interest. .The balance becomes: \(\displaystyle (1.07)t_{n-1}\)
. . Then $100 is withdrawn. .The year-end balance is: .\(\displaystyle (1.07)t_{n-1} - 100\)
This is the balance at the start of the next year.

Therefore: .\(\displaystyle B)\;t_n \;=\;(1.07)t_{n-1}-100\)

 

[margie]

I THANK YOU ALL VERY MUCH FOR YOUR HELP. I TRULY APPRECIATE IT.

U GUYS ARE ALL AWESOME.